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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 242959 Accepted Submission(s):
46863
Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110
注意给数组清零
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main()
{
int m,j,n,i,l1,l2,t,k=1;
char s1[1100];
char s2[1100];
int s3[1100];
int s4[1100];
scanf("%d",&n);
while(n--)
{
scanf("%s %s",s1,s2);
memset(s3,0,sizeof(s3)); //给数组清0
memset(s4,0,sizeof(s4));
l1=strlen(s1);
l2=strlen(s2);
t=0;
for(j=l1-1,i=0;j>=0;j--,i++)
s3[i]=s1[j]-‘0‘; //将字符串转换为数字
for(j=l2-1,i=0;j>=0;i++,j--)
s4[i]=s2[j]-‘0‘;
for(i=0;i<1100;i++)
{
s3[i]+=s4[i];
if(s3[i]>=10) //对各位求和
{
s3[i]-=10;
s3[i+1]++;
}
}
printf("Case %d:\n",k++);
printf("%s + %s = ",s1,s2);
for(i=1099;i>=0;i--)
if(s3[i]!=0)
break;
for(;i>=0;i--)
printf("%d",s3[i]);
printf("\n");
if(n>0)
printf("\n");
}
return 0;
}
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原文地址:http://www.cnblogs.com/tonghao/p/4356781.html